A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression of the solvent if the freezing point constant is -1.86 °C/m?

Answer 1

delta T = Kf*m
Kf = 1.86
m = mols/kg solvent and mols = grams/molar mass. Substitute and solve for m.
Then substitute into dT = KfM and solve for dT. Post your work if you get stuck.